Question

A limnologist wishes to estimate the mean phosphate content per unit volume of lake water. It is knownfrom studies in previous years that the standard deviation has a fairly stable value ofσ= 4. How manyindependent water samples must the limnologist analyze to be 90% certain that the error of estimation [halfwidth of 90% confidence interval] does not exceed 0.8 milligrams?

Answer 1

Answer: 68

Explanation:

As per given , we have

Population standard deviation :
`\sigma=4`

Critical value for 90% confidence interval =
`z_(\alpha/2)=1.645`

Margin of error : E= 0.8 milligrams

The formula to find the sample size is given by :-

`n=((z_(\alpha/2)\ \sigma)/(E))^2`

i.e.
`n=(((1.645)(4))/(0.8))^2=(8.225)^2`

`=67.650625\approx68`

Hence, the limnologist must analyze 68 samples.